What is the equilibrium expression for the reaction `P_(4(s)) + 5O_(2(g))hArrP_(4)O_(10(s))` ?

Options

A. `K_(c)=([P_(4)O_(10)])/([P_(4)][O_(2)]^(5))`
B. `K_(c)=(1)/([O_(2)]^(5))`
C. `K_(c)=[O_(2)]^(5)`
D. `K_(c)=([P_(4)O_(10)])/(5[P_(4)][O_(2)])`

Answer

B

Explanation

In the heterogeneous equilibrium
`P_(4)(s)+5O_(2)(g) hArr P_(4) O_(10)(s)`
`K_(c)=([P_(4)O_(10)(s)])/([P_(4)(s)][O_(2)(g)]^(5))`
Here `[P_(4)O_(10)(s)]=[P_(4)(s)]=1` ( By convention )
`:. K_(c ) =(1)/([O_(2)]^(5))`